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Recently I bumped into a $7^{th}$-grade problem. Shamefully I can't find any elementary solution for it.
The problem is as follows:
There is given a rectangle $ABCD$ with shorter sides $AD=BC$. Let $BCE$, $ABF~$ be two equilateral triangles with $E~$ inside $ABF$. We are being asked to prove that $DEF$ is a equilateral triangle as well.
Any clue how to defeat it $7th$-graders with elementary methods?
We have $AD = EC,$ $AF = CD,$ and $\angle DAF = 30^\circ = \angle ECD.$ By the side-angle-side property we therefore have congruent triangles $\triangle DAF \cong \triangle ECD.$ Therefore $DF = DE.$
To finish the proof we just need to show that $\angle EDF = 60^\circ,$ which I suppose we can angle-chase in a few ways. One way is, since the sum of angles of a triangle is $180^\circ,$ we have $$\angle CDE + \angle CED = 180^\circ - \angle DCE = 180^\circ - 30^\circ = 150^\circ.$$
Then \begin{align} \angle EDF &= \angle EDC + \angle CDF \\ &= \angle EDC + (\angle ADF - \angle ADC) \\ &= (\angle EDC + \angle ADF) - 90^\circ \\ &= (\angle EDC + \angle CED ) - 90^\circ \\ &= 150^\circ - 90^\circ \\ &= 60^\circ. \\ \end{align}
I'm not sure if this would count as 7th grader mathematics but, there is in fact a way to prove this using ideas strictly from elementary geometry as I'll explain below:
1.) Connect point $F$ with $C$ via segment $FC$. Since $\angle ABF=60$, we know that $\angle FBC=30$ and that implies $\angle FBE=30$ as well. This proves that segment $FB$ is the perpendicular bisector of segment $EC$ and the angle bisector of $\triangle BEC$. It follows that $\triangle FBE$ and $\triangle FBC$ are congruent, therefore $FC=FE$.
2.) Since we know that $\angle FBC=30$, we can also conclude that $\angle FAD=30$ as well. We can also notice that $\triangle FBC$ and $\triangle FAD$ are also congruent via the SAS property, therefore we can also infer that $FC=FE=FD$. This proves that point $F$ is the circumcenter of $\triangle DEC$, therefore $\angle DFE$ would be twice the measure of $\angle ECD$. We can see that $\angle ECD=90-60=30$, therefore $\angle DFE=60$ and since $FE=FD$, this proves that $\triangle DFE$ is in fact equilateral.
Another way. Let the shorter side be $2a$ and the other be $2b$. Referred to cartesian coordinates we can made $$A=(0,2a),B=(2b,2a),C=(2b,0),D=(0,0)$$ so we have because of equilateral triangles $\triangle{BEC}$ and $\triangle{AFB}$ $$E=(2b-a\sqrt3,a) \text { and } F=(b,2a-b\sqrt3)$$ With this we have immediately $$\overline{ED}=\overline{DF}=\overline{EF}=2\sqrt{a^2+b^2-ab\sqrt3}$$
I think I've found a good 7th-grade explanation with no equations needed. It requires drawing another equilateral triangle connecting points A and D and then comparing angles.
We're given that triangle AFB is equilateral. Thus, angle AFB is 60 degrees.
Notice that line BF is a perpendicular bisector of line EC. I've labeled the intersection point M in the illustration.
Because of this, FC = FE = FD.
Angle DFE must therefore also be 60 degrees, because angles DFA and EFB are equal as you can see in the diagram (triangles EFM and DFN are congruent).
Since angle DFE is 60 degrees, and DF = FE, triangle DFE must be equilateral.
Here is an approach that aims to show $DE = EF = FD$ to show that $\Delta DEF$ is equilateral.
Firstly, as $DA = CB$, $AF = BF$, and $\angle DAF = \angle CBF = 90º - 60º$, $\Delta DAF$ is congruent to $\Delta CBF$ and so $FD = FC$. Now $\angle ABE = \angle FBC = 90º - 60º$ and thus $\angle ABE = \angle EBF = \angle FBC = 30º$. By $SAS$ congruence again, $\Delta FEB$ is congruent to $\Delta FCB$, and hence $FE = FC$, which yields $FD = FE = FC$.
Using $SAS$ again on $\Delta AEB$ and $\Delta DEC$ (or the fact that $E$ lies on the perpendicular bisector of $BC$), it is not too hard to see that $AE = ED$. However as mentioned previously, $\angle ABE = \angle EBF$, and once more using $SAS$, $AE = EF$.
Thus $AE = EF = ED$ from above, and from before, $FE = FD$.
Therefore $EF = ED = FD$, hence proved.
